[LintCode] 378 Convert Binary Search Tree to Doubly Linked List 解题报告

Description
Convert a binary search tree to doubly linked list with in-order traversal.



Example
Given a binary search tree:

    4
   / \
  2   5
 / \
1   3
return 1<->2<->3<->4<->5

5/11/2017

算法班

直接iterative in order traversal

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Definition for Doubly-ListNode.
 * public class DoublyListNode {
 *     int val;
 *     DoublyListNode next, prev;
 *     DoublyListNode(int val) {
 *         this.val = val;
 *         this.next = this.prev = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param root: The root of tree
     * @return: the head of doubly list node
     */
    public DoublyListNode bstToDoublyList(TreeNode root) {  
        // Write your code here
        if (root == null) return null;

        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;
        DoublyListNode head = null;
        DoublyListNode prev = null;
        
        while (cur != null || !stack.empty()) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            DoublyListNode node = new DoublyListNode(cur.val);
            if (head == null) {
                head = node;
            }
            node.prev = prev;
            if (prev != null) {
                prev.next = node;
            }
            cur = cur.right;
            prev = node;
        }
        return head;
    }
}